1. **State the problem:** We have points A(1, -2) and B on a circle with center P. M(3, 1) is the midpoint of AB. Line l passes through M and P. We need to find: (a) The slope of line AB. (b) The equation of line l. (c) Given the x-coordinate of P is 6, find the y-coordinate of P using the equation of l. 2. **Find the slope of AB:** The midpoint M of AB is given by $$M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$$ Given $M = (3,1)$ and $A = (1,-2)$, let $B = (x_B, y_B)$. From the midpoint formula: $$3 = \frac{1 + x_B}{2} \implies 6 = 1 + x_B \implies x_B = 5$$ $$1 = \frac{-2 + y_B}{2} \implies 2 = -2 + y_B \implies y_B = 4$$ So, $B = (5,4)$. Slope of AB is: $$m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{4 - (-2)}{5 - 1} = \frac{6}{4} = \frac{3}{2}$$ 3. **Find the equation of line l passing through M and P:** Let $P = (6, y_P)$. Slope of line l is: $$m_l = \frac{y_P - 1}{6 - 3} = \frac{y_P - 1}{3}$$ Since P is the center of the circle and A and B lie on the circle, the radius $PA$ is perpendicular to chord AB. Slope of radius $PA$ is: $$m_{PA} = \frac{y_P - (-2)}{6 - 1} = \frac{y_P + 2}{5}$$ Because radius $PA$ is perpendicular to chord $AB$, their slopes satisfy: $$m_{PA} \times m_{AB} = -1$$ Substitute values: $$\left(\frac{y_P + 2}{5}\right) \times \frac{3}{2} = -1$$ Multiply both sides by 10: $$3(y_P + 2) = -10$$ Simplify: $$3y_P + 6 = -10$$ $$3y_P = -16$$ $$y_P = -\frac{16}{3}$$ 4. **Equation of line l:** Slope of l is: $$m_l = \frac{y_P - 1}{3} = \frac{-\frac{16}{3} - 1}{3} = \frac{-\frac{16}{3} - \frac{3}{3}}{3} = \frac{-\frac{19}{3}}{3} = -\frac{19}{9}$$ Using point-slope form with point M(3,1): $$y - 1 = -\frac{19}{9}(x - 3)$$ Simplify: $$y = -\frac{19}{9}x + \frac{19}{3} + 1 = -\frac{19}{9}x + \frac{19}{3} + \frac{3}{3} = -\frac{19}{9}x + \frac{22}{3}$$ 5. **Verify y-coordinate of P using line l:** Substitute $x=6$: $$y = -\frac{19}{9} \times 6 + \frac{22}{3} = -\frac{114}{9} + \frac{22}{3} = -\frac{38}{3} + \frac{22}{3} = -\frac{16}{3}$$ This matches the $y_P$ found earlier. **Final answers:** (a) Slope of AB is $\frac{3}{2}$. (b) Equation of line l is $$y = -\frac{19}{9}x + \frac{22}{3}$$. (c) The y-coordinate of P is $-\frac{16}{3}$ (approximately -5.33).